Hướng dẫn:

$\begin{align}

  & N{{a}_{2}}O+{{H}_{2}}O\to 2NaOH\left( 1 \right) \\

 & 2NaOH+{{H}_{2}}S{{O}_{4}}\to N{{a}_{2}}S{{O}_{4}}+{{H}_{2}}O\left( 2 \right) \\

\end{align}$

${{n}_{N{{a}_{2}}O}}=\frac{{{m}_{N{{a}_{2}}O}}}{{{M}_{N{{a}_{2}}O}}}=\frac{6,2}{62}=0,1$

Theo phương trình (1): ${{n}_{NaOH}}=2{{n}_{N{{a}_{2}}O}}=0,1.2=0,2$ (mol)

Theo phương trình (2): ${{n}_{{{H}_{2}}S{{O}_{4}}}}=\frac{1}{2}{{n}_{NaOH}}=\frac{1}{2}.0,2=0,1$(mol)

${{V}_{{{H}_{2}}S{{O}_{4}}}}=\frac{{{n}_{{{H}_{2}}S{{O}_{4}}}}}{{{C}_{{{M}_{{{H}_{2}}S{{O}_{4}}}}}}}=\frac{0,1}{0,1}=1$ (lít)

                $\Rightarrow $ Chọn đáp án B.